Question

What is the square root of -2i?

Answer 1

1-i and -1+i

Explanation:

We are to find the square roots of
`z=0-2i`. First, convert from Cartesian to polar form:

`r=√(a^2+b^2)\\r=√(0^2+(-2)^2)\\r=√(0+4)\\r=√(4)\\r=2`

`\theta=tan^(-1)((b)/(a))\\ \theta=tan^(-1)((-2)/(0))\\\theta=(3\pi)/(2)`

`z=2(\cos(3\pi)/(2)+i\sin(3\pi)/(2))`

Next, use the formula
`\displaystyle \sqrt[n]{r}\biggr[\cis\biggr((\theta+2\pi k)/(n)\biggr)\biggr]` where
`\displaystyle k=0,1,2,...\:,n-1` to find the square roots:

When k=1

`\displaystyle \sqrt[2]{2}\biggr[cis\biggr(((3\pi)/(2)+2\pi(1))/(2)\biggr)\biggr]`

`\displaystyle √(2)\biggr[cis\biggr((3\pi)/(4)+\pi\biggr)\biggr]`

`√(2)\biggr(cis(7\pi)/(4)\biggr)`

`√(2)(\cos(7\pi)/(4)+i\sin(7\pi)/(4))\\ \\√(2)((√(2))/(2)-(√(2))/(2)i)\\ \\1-i`

When k=0

`\displaystyle \sqrt[2]{2}\biggr[cis\biggr(((3\pi)/(2)+2\pi(0))/(2)\biggr)\biggr]`

`√(2)\biggr(cis(3\pi)/(4)\biggr)`

`√(2)(\cos(3\pi)/(4)+i\sin(3\pi)/(4))\\ \\√(2)(-(√(2))/(2)+(√(2))/(2)i)\\ \\-1+i`

Thus, the square roots of -2i are 1-i and -1+i